p(x) be polynomial of dgree atmost 5

which leaves remainder -1 and +1

upon division by (x-1)^3 and (x+1)^3 respectively.

Q1-no. of real roots of p(x):

a)1               b)2                c)3               d)5

Q2-the max value of y=p' '(x) can be obtained at x=

a) -1/root3            b)0            c) 1/root3            d)1

Q3-the sum of pairwise product of all

the roots (real and complex) of p(x)=0 is:

a) -5/3                b)-10/3           c)2               d) -5

-ROHIT KUMAR RAJPOOT

 p(x)=(Ax² + Bx + C)(x+1)³ + 1 ……………………..<i>

p(x)=(Dx² + Ex +F)(x-1)³ – 1 ………………………….<ii>

Equating <i> & <ii> , we get 6 equations:- A=D  ;  C+1=-F-1  ;  3A+B=E-3D  ;  3A+3B+C=3D-3E+F  ; 

3E-D-3F=A+3B+3C  ;  3F-E=B+3C

Solving, we get    A=-3/8        ;    B=9/8    ;    C=-1    ;    D=-3/8   ;    E=-9/8    ;    F=-1

Hence     p(x)=(-3/8)x⁵ +(5/4) x³ +(-15/8)x

p(x) has only one real root x=0

p”(x)=(-15/2)(x³ – x) has a maximum at x=1/√3